Integrand size = 106, antiderivative size = 24 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=e^{-e^{4 x (x+\log (-x+\log (3)))}-x} x \] Output:
x/exp(x+exp(4*x*(ln(ln(3)-x)+x)))
\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx \] Input:
Integrate[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*L og[3] + E^(4*x^2 + 4*x*Log[-x + Log[3]])*(4*x^2 + 8*x^3 - 8*x^2*Log[3] + ( 4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]),x]
Output:
Integrate[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*L og[3] + E^(4*x^2 + 4*x*Log[-x + Log[3]])*(4*x^2 + 8*x^3 - 8*x^2*Log[3] + ( 4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]), x]
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-e^{4 x^2+4 x \log (\log (3)-x)}-x} \left (x^2+e^{4 x^2+4 x \log (\log (3)-x)} \left (8 x^3+4 x^2-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (\log (3)-x)\right )-x+(1-x) \log (3)\right )}{\log (3)-x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (4 x e^{4 x^2-e^{4 x^2+4 x \log (\log (3)-x)}-x} \left (2 x^2+x \log (\log (3)-x)+x (1-\log (9))-\log (3) \log (\log (3)-x)\right ) (\log (3)-x)^{4 x-1}+e^{-e^{4 x^2+4 x \log (\log (3)-x)}-x}-x e^{-e^{4 x^2+4 x \log (\log (3)-x)}-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{-x-e^{4 x^2+4 \log (\log (3)-x) x}}dx-\int e^{-x-e^{4 x^2+4 \log (\log (3)-x) x}} xdx+4 (1-\log (9)) \int e^{4 x^2-x-e^{4 x^2+4 \log (\log (3)-x) x}} x^2 (\log (3)-x)^{4 x-1}dx-4 \int e^{4 x^2-x-e^{4 x^2+4 \log (\log (3)-x) x}} x (\log (3)-x)^{4 x} \log (\log (3)-x)dx+8 \int e^{4 x^2-x-e^{4 x^2+4 \log (\log (3)-x) x}} x^3 (\log (3)-x)^{4 x-1}dx\) |
Input:
Int[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*Log[3] + E^(4*x^2 + 4*x*Log[-x + Log[3]])*(4*x^2 + 8*x^3 - 8*x^2*Log[3] + (4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]),x]
Output:
$Aborted
Time = 1.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(x \,{\mathrm e}^{-x -{\mathrm e}^{4 x \left (\ln \left (\ln \left (3\right )-x \right )+x \right )}}\) | \(21\) |
risch | \(x \,{\mathrm e}^{-\left (\ln \left (3\right )-x \right )^{4 x} {\mathrm e}^{4 x^{2}}-x}\) | \(26\) |
Input:
int((((-4*x*ln(3)+4*x^2)*ln(ln(3)-x)-8*x^2*ln(3)+8*x^3+4*x^2)*exp(4*x*ln(l n(3)-x)+4*x^2)+(1-x)*ln(3)+x^2-x)/(ln(3)-x)/exp(exp(4*x*ln(ln(3)-x)+4*x^2) +x),x,method=_RETURNVERBOSE)
Output:
x/exp(x+exp(4*x*(ln(ln(3)-x)+x)))
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \] Input:
integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*ex p(4*x*log(log(3)-x)+4*x^2)+(1-x)*log(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log( log(3)-x)+4*x^2)+x),x, algorithm="fricas")
Output:
x*e^(-x - e^(4*x^2 + 4*x*log(-x + log(3))))
Timed out. \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\text {Timed out} \] Input:
integrate((((-4*x*ln(3)+4*x**2)*ln(ln(3)-x)-8*x**2*ln(3)+8*x**3+4*x**2)*ex p(4*x*ln(ln(3)-x)+4*x**2)+(1-x)*ln(3)+x**2-x)/(ln(3)-x)/exp(exp(4*x*ln(ln( 3)-x)+4*x**2)+x),x)
Output:
Timed out
Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \] Input:
integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*ex p(4*x*log(log(3)-x)+4*x^2)+(1-x)*log(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log( log(3)-x)+4*x^2)+x),x, algorithm="maxima")
Output:
x*e^(-x - e^(4*x^2 + 4*x*log(-x + log(3))))
\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int { -\frac {{\left (x^{2} + 4 \, {\left (2 \, x^{3} - 2 \, x^{2} \log \left (3\right ) + x^{2} + {\left (x^{2} - x \log \left (3\right )\right )} \log \left (-x + \log \left (3\right )\right )\right )} e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )} - {\left (x - 1\right )} \log \left (3\right ) - x\right )} e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )}}{x - \log \left (3\right )} \,d x } \] Input:
integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*ex p(4*x*log(log(3)-x)+4*x^2)+(1-x)*log(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log( log(3)-x)+4*x^2)+x),x, algorithm="giac")
Output:
integrate(-(x^2 + 4*(2*x^3 - 2*x^2*log(3) + x^2 + (x^2 - x*log(3))*log(-x + log(3)))*e^(4*x^2 + 4*x*log(-x + log(3))) - (x - 1)*log(3) - x)*e^(-x - e^(4*x^2 + 4*x*log(-x + log(3))))/(x - log(3)), x)
Time = 3.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x^2}\,{\left (\ln \left (3\right )-x\right )}^{4\,x}}\,{\mathrm {e}}^{-x} \] Input:
int((exp(- x - exp(4*x^2 + 4*x*log(log(3) - x)))*(x + log(3)*(x - 1) - x^2 + exp(4*x^2 + 4*x*log(log(3) - x))*(log(log(3) - x)*(4*x*log(3) - 4*x^2) + 8*x^2*log(3) - 4*x^2 - 8*x^3)))/(x - log(3)),x)
Output:
x*exp(-exp(4*x^2)*(log(3) - x)^(4*x))*exp(-x)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\frac {x}{e^{e^{4 x^{2}} \left (\mathrm {log}\left (3\right )-x \right )^{4 x}+x}} \] Input:
int((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*exp(4*x* log(log(3)-x)+4*x^2)+(1-x)*log(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log(log(3) -x)+4*x^2)+x),x)
Output:
x/e**(e**(4*x**2)*(log(3) - x)**(4*x) + x)